Re: Lightning and Fourier transform of an impulse

On Aug 7, 11:44 am, cs_post...@xxxxxxxxxxx wrote
(among other things) that

Or to put it bluntly, the frequency domain is practically speaking a
bad way to represent one-shot or random (non-periodic) events.

ALL real-life signals are one-shot and nonperiodic. NO human being
has observed an alleged periodic signal, say sin(wt), for ALL t to
verify
that it indeed satisfies the conditions that a periodic signal must
satisfy. For all you know, the signal is going to disappear at 5 pm
when someone turns off the oscillator and goes home. So let's just
trash the frequency domain and forget the notion entirely.

The Fourier transform of the finite-duration (i.e. one-shot) pulse
rect(t) is sinc(f). What this says in terms of the Fourier integral
(or inverse Fourier transform) is that if we have (uncountably many)
oscillators (at all possible frequencies) with infinitesimally small
amplitudes (given by sinc(f)) and all in phase (peaking at t = 0),
then
these complex exponential signals, lasting for all time, have the
curious property that their "sum" cancels out and results in 0 at
all time instants t such that |t| > 1/2, while for |t| < 1/2, these
signals
"add up" to 1 exactly. We accept this notion (or I like to think that
many denizens of comp.dsp do) and use it all the time.

The pulse rect (t) delivers energy to a load only during the interval
(-1/2, 1/2). The complex oscillators are running for all t, but
since
each has infinitesimal amplitude, each individual oscillator delivers
zero energy. To get any energy from them, we have to use an
uncountable number. If we use all the oscillators, then since the
"sum"
disappears for |t| > 1/2, there is no energy delivered for |t| > 1/2.
If we use a subset of them (say all the oscillators whose frequencies
are less than 10 Hz) by passing rect(t) through a lowpass filter,
then
energy *will* be delivered to the filter load even after t = 1/2, and
indeed even before t = -1/2 if one thinks of ideal LPFs which are
noncausal filters.

Just to muddy the waters even further....
.

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